How do you find the axis of symmetry, and the maximum or minimum value of the function #f(x)= 4x^2+40x+97#?

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Answer 1 · 2016-05-20T07:20:39

Axis of symmetry is #x=-5# and minima at #(-5,-3)#

For an equation of a parabola given by the equation

#y=ax^2+bx+c#, the axis of symmetry is a vertical line given by #x=-b/2a#

Hence, axis of symmetry for #y=4x^2+40x+97# is #x=-40/(2xx4)=-5#

As the differential #(dy)/(dx)=8x+40# and this is zero at #8x+40=0# or #x=-5#. At this value #y=4(-5)x^2+40(-5)+97=100-200+97=-3#

As second derivative #(d^2y)/dx^2)=8# and is positive

hence we have a miniima at #(-5,-3)#

graph{4x^2+40x+97 [-7.52, -2.52, -3.53, -1.03]}